Question

In: Statistics and Probability

(1 point) Using the Standard Normal Table found in your textbook, find the z-scores such that:...

(1 point) Using the Standard Normal Table found in your textbook, find the z-scores such that:

(a) The area under the standard normal curve to its left is 0.5

z =

(b) The area under the standard normal curve to its left is 0.9826

z =

(c) The area under the standard normal curve to its right is 0.1423

z =

(d) The area under the standard normal curve to its right is 0.9394

z =

Expert Solution

(a) The area under the standard normal curve to its left is 0.5

P(Z<x) = 0.5

From z table P = 0.5 obtained at z = 0

So Z score = 0

(b) The area under the standard normal curve to its left is 0.9826

P(Z<x) = 0.9826

From Z table p = 0.9826 Obtained z = 2.111

Z score = 2.111

(c) The area under the standard normal curve to its right is 0.1423

P(Z>x) = 0.1423

P(Z>x) = 1- P(Z<x) = 0.1423

1- P(Z<x) = 0.1423

P(Z<x) = 1-0.1423

P(Z<x) = 0.8577

From z table p = 0.8577 obtained at z = 1.07

Z score = 1.07

(d) The area under the standard normal curve to its right is 0.9394

P(Z>x) = 0.9394

P(Z>x) = 1- P(Z<x) = 0.9394

1- P(Z<x) = 0.9394

P(Z<x) = 1-0.9394

P(Z<x) = 0.0606

From z table p = 0.0606 obtained at z = -1.55

Z score = -1.55

orchestra answered 3 years ago

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